Answer
$\pm 1;\pm \dfrac {1}{2};\pm \dfrac {1}{4};\pm 2;\pm 3;\pm \dfrac {3}{2};\pm \dfrac {3}{4};\pm 6$
Work Step by Step
In a polinomial function like
$f\left( x\right) =a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots a_{1}x+a_{0}$
If $p/q$, in lowest terms, is a rational zero of $f$, then p must be a factor of $a_0 $ and $q$ must be a factor of $a_n$.
Here
$f\left( x\right) =-4x^{3}+x^{2}+x+6\Rightarrow a_{n}=-4;a_{0}=6$
Factors of $a_0$ are $\pm 1,\pm 2,\pm 3,\pm 6$
Factors of $a_n$ are $\pm 1,\pm 2,\pm 4$
So the potential rational zeros are:
$\pm 1;\pm \dfrac {1}{2};\pm \dfrac {1}{4};\pm 2;\pm 3;\pm \dfrac {3}{2};\pm \dfrac {3}{4};\pm 6 $